We can then perform quantifier elimination. Note that this is not possible in case we do not have the $card$ operator : \[\exists v.card(S_1 \cap v) = k_1 \wedge card(S_1 \cap v') = k_2 \Rightarrow card(S_1) = k_1 + k_2\]

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We can then perform quantifier elimination. Note that this is not possible in case we do not have the $card$ operator : \begin{equation*}\exists v.card(S_1 \cap v) = k_1 \wedge card(S_1 \cap v') = k_2 \Rightarrow card(S_1) = k_1 + k_2\]

After elimination of the quantifiers,​ deciding the result of the formula is trivial.

After elimination of the quantifiers,​ deciding the result of the formula is trivial.

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If we add to our language the possibility ot replace $k$ by $card(S)$ then this new language is still decidable and is equivalent to Presbulgarian arithmetic !